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Solve-dy-dx-y-tan-x-y-2-tan-2-x-please-help-




Question Number 5908 by sanusihammed last updated on 04/Jun/16
Solve    (dy/dx)  − y tan x  = y^2  tan^2  x    please help.
$${Solve}\:\: \\ $$$$\frac{{dy}}{{dx}}\:\:−\:{y}\:{tan}\:{x}\:\:=\:{y}^{\mathrm{2}} \:{tan}^{\mathrm{2}} \:{x} \\ $$$$ \\ $$$${please}\:{help}. \\ $$
Commented by 123456 last updated on 05/Jun/16
(dy/dx)=ytan x+y^2 tan^2 x  (dy/dx)=ytan x(1+ytan x)
$$\frac{{dy}}{{dx}}={y}\mathrm{tan}\:{x}+{y}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} {x} \\ $$$$\frac{{dy}}{{dx}}={y}\mathrm{tan}\:{x}\left(\mathrm{1}+{y}\mathrm{tan}\:{x}\right) \\ $$
Commented by nburiburu last updated on 25/Jun/16
it is a Bernoulli diff. eq. and can be solved with z=y^(−1)  transforming to a linear diff. eq. of form y′+P y=Q which has integration factor μ=e^(∫Pdx)  and μ.y=∫μ.Qdx
$${it}\:{is}\:{a}\:{Bernoulli}\:{diff}.\:{eq}.\:{and}\:{can}\:{be}\:{solved}\:{with}\:{z}={y}^{−\mathrm{1}} \:{transforming}\:{to}\:{a}\:{linear}\:{diff}.\:{eq}.\:{of}\:{form}\:{y}'+{P}\:{y}={Q}\:{which}\:{has}\:{integration}\:{factor}\:\mu={e}^{\int{Pdx}} \:{and}\:\mu.{y}=\int\mu.{Qdx} \\ $$

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